Convex cyclic hexagon $ABCDEF$. Prove $AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA$ 
Convex hexagon $ABCDEF$ inscribed within a circle. Prove that
$$AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA\,.$$

I was thinking of represending the inequalities in trigonometry then use Language multiplier. For example let $\angle AOB = \theta_1$, $\angle BOC = \theta_2$, represent the inequality in trigonometry, subject to constraint $\theta_1 + \theta_2 + ... + \theta_6 = 2\pi$. But it's still quite a bit of work and I didn't manage to get to the end. It also seems a bit overkill -- might be better solution? Would like to see any approach.
 A: Nice problem!
Let me post a solution using inversion and cross-ratios.
One can rewrite the inequality in the following way:
$$(ABCF)\cdot(BCDA)\cdot(CDEB)\cdot(DEFC)\cdot(EFAD)\cdot(FABE)\ge 729 \qquad (\heartsuit)$$
where for brevity we write $(XYZT)$ for the crossratio $(X,Y;Z,T)$.
Consider an inversion with respect to a circle centered at $F$. Let the images of $A,B,C,D,E$ be $A', B', C', D', E'$, respectively. By basic properties of inversion these points lie on a common line, say $\ell$. Denote the point at infinity of $\ell$ by $F'$. For every quadruple $X,Y,Z,T$ such that $(XYZT)$ appears in $(\heartsuit)$ we have $(XYZT)=(FX,FY;FZ,FT)=(FX',FY';FZ',FT')=(X'Y'Z'T')$ where $FF$ is understood as the line tangent to the circumcircle of $ABCDEF$ at $F$. Therefore we have to prove a variant of $(\heartsuit)$ in which every letter $X$ is replaced by $X'$; call the new inequality $(\spadesuit)$.
Since $ABCDEF$ is convex, points $A',B',C',D',E'$ lie on $\ell$ in this order. Denote $2x=A'B', y=B'C', z=C'D', 2t=D'E'$. Then $(\spadesuit)$ can be written as
$$\frac{(z+2t)(y+z)(2x+y)(2x+y+z+2t)}{xyzt}\ge 108.$$
This follows from AM-GM: just multiply the following:
\begin{align*}
z+2t &\ge 3z^{1/3}t^{2/3}, \\
y+z &\ge 2y^{1/2}z^{1/2}, \\
2x+y &\ge 3x^{2/3}y^{1/3}, \\
2x+y+z+2t &\ge 6x^{2/6}y^{1/6}z^{1/6}t^{2/6}.
\end{align*}
A: This problem had been posted here, but was deleted by the owner for an unspecified reason.  I flagged the moderators about this, but they didn't do anything.  Here is the same solution I gave in that link.
Ptolemy's theorem with $\square ABCD$ yields
$$AB\cdot CD+AD\cdot BC=AC\cdot BD.$$
Ptolemy's theorem with $\square ACDE$ yields
$$AC\cdot DE+EA\cdot CD=AD\cdot CE.$$
Hence,
$$AD=\frac{AC\cdot DE+EA\cdot CD}{CE}$$
so that
$$AC\cdot BD=AB\cdot CD+AD\cdot BC=AB\cdot CD+\left(\frac{AC\cdot DE+EA\cdot CD}{CE}\right)\cdot BC.$$
Hence
$$AC\cdot BD=AB\cdot CD+\frac{AC}{CE}(BC\cdot DE)+\frac{EA}{CE}(BC\cdot CD).$$
By AM-GM,
$$AC\cdot BD\geq 3\sqrt[3]{(AB\cdot CD)\left(\frac{AC}{CE}(BC\cdot DE)\right)\left(\frac{EA}{CE}(BC\cdot CD)\right)}=3\sqrt[3]{AB\cdot BC^2\cdot  CD^2\cdot DE\cdot \frac{AC\cdot EA}{CE^2}}.$$
This shows that
$$\sqrt[3]{\frac{AC^2\cdot BD^3\cdot CE^2}{EA}}\geq 3\sqrt[3]{AB\cdot BC^2\cdot CD^2\cdot DE}.$$
Similarly,
$$\sqrt[3]{\frac{BD^2\cdot CE^3\cdot DF^2}{FB}}\geq 3\sqrt[3]{BC\cdot CD^2\cdot DE^2\cdot EF},$$
$$\sqrt[3]{\frac{CE^2\cdot DF^3\cdot EA^2}{AC}}\geq 3\sqrt[3]{CD\cdot DE^2\cdot EF^2\cdot FA},$$
$$\sqrt[3]{\frac{DF^2\cdot EA^3\cdot FB^2}{BD}}\geq 3\sqrt[3]{DE\cdot EF^2\cdot FA^2\cdot AB},$$
$$\sqrt[3]{\frac{EA^2\cdot FB^3\cdot AC^2}{CE}}\geq 3\sqrt[3]{EF\cdot FA^2\cdot AB^2\cdot BC},$$
and
$$\sqrt[3]{\frac{FB^2\cdot AC^3\cdot BD^2}{DF}}\geq 3\sqrt[3]{FA\cdot AB^2\cdot BC^2\cdot CD}.$$
Multiplying all six inequalities above gives
$$(AC\cdot BD\cdot CE\cdot DF\cdot EA\cdot FB)^2\geq (27\cdot AB\cdot BC\cdot DE\cdot EF\cdot FA)^2\,,$$
which is equivalent to the required inequality.  The equality holds if and only if $ABCDEF$ is a regular hexagon.
